Introduction

In this blog post, we will explore how to solve the LeetCode problem "718" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays. Example 1: Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7] Output: 3 Explanation: The repeated subarray with maximum length is [3,2,1]. Example 2: Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0] Output: 5 Explanation: The repeated subarray with maximum length is [0,0,0,0,0]. Constraints: 1 <= nums1.length, nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 100

Explanation

Here's an efficient solution to find the maximum length of a common subarray between two arrays:

• Dynamic Programming: Utilize a 2D DP table where dp[i][j] stores the length of the longest common subarray ending at nums1[i-1] and nums2[j-1].
• State Transition: If nums1[i-1] == nums2[j-1], then dp[i][j] = dp[i-1][j-1] + 1. Otherwise, dp[i][j] = 0.

• Maximize: Keep track of the maximum value in the DP table, which represents the maximum length of the common subarray.

• Time Complexity: O(m*n), where m and n are the lengths of nums1 and nums2 respectively.

• Space Complexity: O(m*n)

Code

    def findLength(nums1, nums2):
    """
    Finds the maximum length of a subarray that appears in both arrays.

    Args:
        nums1: The first integer array.
        nums2: The second integer array.

    Returns:
        The maximum length of a common subarray.
    """

    n = len(nums1)
    m = len(nums2)
    dp = [[0] * (m + 1) for _ in range(n + 1)]
    max_length = 0

    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if nums1[i - 1] == nums2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
                max_length = max(max_length, dp[i][j])
            else:
                dp[i][j] = 0

    return max_length